Lines and Angles Test

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Lines and Angles Test - 35

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Weekly Quiz Competition

Question 1
1 / -0

In the given figure, $$\angle QPB$$ is

A
$${ 60 }^{ o }$$

B
$${ 45 }^{ o }$$

C
$${ 30 }^{ o }$$

D
$${ 15 }^{ o }$$

Solution

$$2x+3x+x=180^o(\text{linear pair})$$
$$6x=180^o$$
$$x=30^o$$
Question 2
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Lines $$PQ$$ and $$RS$$ intersect at $$O$$. If $$\angle POS=2\angle SOQ$$ then the four angles at $$O$$ are

A
$${ 30 }^{ o },{ 30 }^{ o },{ 120 }^{ o },{ 180 }^{ o }$$

B
$${ 60 }^{ o },{ 60 }^{ o },{ 120 }^{ o },{ 120 }^{ o }$$

C
$${ 60 }^{ o },{ 90 }^{ o },{ 90 }^{ o },{ 120 }^{ o }$$

D
$${ 30 }^{ o },{ 60 }^{ o },{ 90 }^{ o },{ 180 }^{ o }$$

Solution

$$x+2x=180 (linear pair)\Rightarrow x={ 60 }^{ o }\Rightarrow \angle SOP={ 120 }^{ o }$$
$$\therefore \text{four angle are at O}\quad are\ { 60 }^{ o },{ 60 }^{ o },{ 120 }^{ o },{ 120 }^{ o }\quad $$
Question 3
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The angle between two opposite rays is ________.

A
Right

B
Obtuse

C
Acute

D
Straight

Solution
Question 4
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Which of the following is an obtuse angle?

A
$$92^{\circ}$$

B
$$181^{\circ}$$

C
$$195^{\circ}$$

D
$$83^{\circ}$$
Question 5
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A transversal intersects two or more than two lines at _________ points.

A
Same

B
Different

C
Equidistant

D
None of these

Solution

A transversal intersects two or more than two lines at different points.
Question 6
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Find the value of x, y and z in the adjoining figure.

A
$$x=160$$, $$y=60$$, $$z=80$$

B
$$x=60$$, $$y=80$$, $$z=40$$

C
$$x=80$$, $$y=60$$, $$z=140$$

D
$$x=140$$, $$y=80$$, $$z=60$$

Solution

$$In\triangle BCE$$
$$\implies\quad \angle BEC+\angle BCE+\angle CBE={ 180 }^{ \circ }$$
$$\implies\quad { 90 }^{ \circ }+{ 30 }^{ \circ }+\angle CBE={ 180 }^{ \circ }$$
$$\implies\quad \angle CBE={ 60 }^{ \circ }$$
$$ y={ 60 }^{ \circ }$$
$$ In\triangle APC:$$
$$\implies\quad { 50 }^{ \circ }+{ 30 }^{ \circ }+\angle APC={ 180 }^{ \circ }$$
$$\implies\quad \angle APC={ 100 }^{ \circ }$$
$$ \therefore x=180-\angle APC$$
$$ (\because sum\quad of\quad angles\quad on\quad a\quad straight\quad line={ 180 }^{ \circ })$$
$$\implies\quad x={ 180 }^{ \circ }-{ 100 }^{ \circ }={ 80 }^{ \circ }$$
$$ In\triangle BOP:$$
$$ z=x+y\quad (exterior\quad angle\quad of\quad a\quad triangle=sum\quad of\quad two\quad opposite\quad interior\quad angles)$$
$$\implies\quad z=80+60={ 140 }^{ \circ }$$
$$ \therefore x={ 80 }^{ \circ }\quad ,\quad y={ 60 }^{ \circ }\quad ,z={ 140 }^{ \circ }$$
Question 7
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In $$\triangle ABC,\angle A={ x }^{ \circ },\angle B={ \left( 2x-15 \right) }^{ \circ}$$ and $$\angle C ={ \left( 3x+21 \right) }^{ \circ }$$. Find the value of $$x$$ and the measure of each angle of the triangle.

A
$$x={ 27 }^{ \circ },\angle A={ 27 }^{ \circ },\angle B={ 41 }^{ \circ },\angle C={ 101 }^{ \circ }$$

B
$$x={ 29 }^{ \circ },\angle A={ 29 }^{ \circ },\angle B={ 43 }^{ \circ },\angle C={ 108 }^{ \circ }$$

C
$$x={ 27 }^{ \circ },\angle A={ 27 }^{ \circ },\angle B={ 41 }^{ \circ },\angle C={ 108 }^{ \circ }$$

D
$$x={ 30 }^{ \circ },\angle A={ 30 }^{ \circ },\angle B={ 41 }^{ \circ },\angle C={ 109 }^{ \circ }$$

Solution

Given, $$\angle A=x^\circ$$, $$\angle B=(2x-15)^\circ$$ and $$\angle C=(3x+21)^\circ$$.
We know, by angle sum property, the sum of angles of a triangle is $$180^\circ$$.
Then, $$\angle A+ \angle B+ \angle C=180^\circ$$
$$\implies$$ $$x^\circ+(2x-15)^\circ+ (3x+21)^\circ=180^\circ$$
$$\implies$$ $$6x^\circ+6^\circ=180^\circ$$
$$\implies$$ $$6x^\circ=180^\circ-6^\circ$$
$$\implies$$ $$6x^\circ=174^\circ$$
$$\implies$$ $$x^\circ=29^\circ$$.

Therefore,
$$\angle A=x^\circ=29^\circ$$,
$$\angle B=(2x-15)^\circ=2\times29^\circ-15^o=58^\circ-15^\circ=43^\circ$$
and $$\angle C=(3x+21)^\circ=3\times29^\circ+21^\circ=87^\circ+21^\circ=108^\circ$$.

Hence, option $$B$$ is correct.
Question 8
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From the figure, find values of $$x$$ and $$y$$.

A
$$25^{o},30^{o}$$

B
$$35^{o},31^{o}$$

C
$$50^{o},28^{o}$$

D
$$45^{o},33^{o}$$

Solution

In the given triangle,
by angle sum property,
$$40^o+95^o+x^o=180^o.......(i)$$
and $$x^o+y^o+102^o=180^o.......(ii)$$.

From $$(i)$$,
$$40^o+95^o+x^o=180^o$$
$$\implies$$ $$135^o+x^o=180^o$$
$$\implies$$ $$x^o=180^o-135^o$$
$$\implies$$ $$x^o=45^o.......(iii)$$.

Substitute $$(iii)$$ in $$(ii)$$,
$$45^o+y^o+102^o=180^o$$
$$\implies$$ $$y^o+147^o=180^o$$
$$\implies$$ $$y^o=180^o-147^o$$
$$\implies$$ $$y^o=33^o$$.

Therefore, $$x^o=45^o$$ and $$y^o=33^o$$.
Hence, option $$D$$ is correct.
Question 9
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From the adjacent figure find the values of $$\angle P R S$$ (in degree).

A
$$85$$

B
$$75$$

C
$$65$$

D
$$55$$

Solution

We know, by exterior angle property, an exterior angle of a triangle is equal to the sum of the two interior opposite angles.

Then, in $$\triangle PQR$$,
$$\angle QPR+\angle PQR = \angle PRS$$
$$\implies$$ $$35^{o}+50^{o} = \angle PRS$$
$$\implies$$ $$\angle PRS = 85^{o}$$.

Therefore, option $$A$$ is correct.
Question 10
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Find the complement of the angle $$63^o$$

A
$${70^ \circ}$$

B
$${27^ \circ}$$

C
$${45^ \circ}$$

D
$${62^ \circ}32'$$

Solution

We know, two angles are complementary when they add up to $$90^o.$$

Given, measure of one complementary angle is $$63^o.$$

$$\Rightarrow$$ Measure of other complementary angle $$=90^o-63^o$$ $$=27^o.$$

$$\therefore$$ Measure of a complementary angle of $$ 63^{o}=27^o$$.
Hence, option $$B$$ is correct.